The Price is Right Bidding Strategies

Question

I am currently reading this article.

If you are unfamiliar with the game, "The Price is Right", at the beginning, four bidders are selected who have to guess the price of some consumer object. Whoever gets closest to the retail value (rounded to the dollar) without going over goes on to play a different pricing game. Play is sequential, so you can imagine bidding last provides an advantage. This article is studying whether or not players behave rationally in this first part of the game, or whether it is bounded by some simpler decision rule.

The paper has a proposition:

Suppose that contestants have rational expectations. Then in equilibrium,

1. The fourth bidder must win at least as often as the third bidder; and the third bidder must win at least as often as either the first or second bidder.
2. The fourth bidder must win at least 1/3 of the time.
3. The first and second bidders together cannot win more than 4/9 of the time.

The appendix then, in its proof, starts the following explanation:

First note that if a bidder believes that his probability of winning exceeds that of another bidder, then since every bidder has rational expectations, this belief must be correct. Thus to show that a bidder wins more often than another bidder, it is sufficient to show that he believes that he wins more often.

Which is fine.

First consider the last bidder. Since his optimal strategy is to pick the interval that he believes gives him the largest probability of winning, he must believe that he does at least as well as any previous bidder.

I do not understand this part of the explanation. Is the logic that the last bidder will bid 1 dollar above whoever it will get the best distribution, and then if someone were to have a better chance of winning than the last bidder, the bidder would simply bid 1 dollar above that person?

Answer 1

Yes. Note that in the paper the authors assume that the action space is continuous, so it is possible to bid in arbitrarily small fractions of a dollar.

Suppose players $i\in\{1,2,3\}$ have bids $b_i$ that, without loss of generality, are ordered $b_1<b_2<b_3$. Write $p$ for the actual price.

If we forget about player four for a moment, then players' probability of winning are

$$\Pr(1\text{ wins})=\Pr(b_1\leq p< b_2)\equiv \pi_1$$ $$\Pr(2\text{ wins})=\Pr(b_2\leq p< b_3)\equiv \pi_2$$ $$\Pr(3\text{ wins})=\Pr(b_3\leq p)\equiv \pi_3.$$

Suppose $\pi_1>\pi_2,\pi_3$. Then the optimal thing for 4 to do is to bid $b_4=b_1+\epsilon$ ($\epsilon$ positive and small). This bis would result in $$\Pr(1\text{ wins})=0$$ $$\Pr(2\text{ wins})=\Pr(b_2\leq p< b_3)$$ $$\Pr(3\text{ wins})=\Pr(b_3< p)$$ $$\Pr(4\text{ wins})=\Pr(b_1\leq p< b_2).$$

One can easilly check that $\pi_1>\pi_2,\pi_3$ implies that $\Pr(4\text{ wins})$ is bigger than any rival winning probability. Likewise, if we had $\pi_2>\pi_1,\pi_3$ then the optimal thing for $4$ to do would be to bid $b_4=b_2+\epsilon$.

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edit: The above reasoning works if we take $\pi_1>\pi_2,\pi_3$ (with strict inequality). But what if the inequality is not strict?

Suppose that $b_1<b_2<b_3$, and that $\pi_1=\pi_2>\pi_3,\Pr(p<b_1)$. Then it seems like $b_4=b_1+\epsilon$ must yield a lower winning probability than that enjoyed by player 2 for any positive $\epsilon$. indeed we would have $$\Pr(b_1+\epsilon\leq p<b_2)<\Pr(b_1\leq p<b_2)=\Pr(b_2\leq p<b_3).$$ Similarly, $b_4=b_2+\epsilon$ will yield a lower winning probability than that enjoyed by player 1 for any positive $\epsilon$.

I think that, in equilibrium, this won't happen so I guess the result is fine. But it makes the proof seem a bit hand-wavey.